Demo 3.5: Amplifier Stages, and Choosing the Operating Point

The previous demonstration ended with a load line and a movable point on it, and with a warning: beta is not a design parameter. It varies threefold between two parts with the same number printed on them, it drifts with temperature, and no data sheet will promise you a value.

Those two facts are in tension, and this demonstration is about the tension. An amplifier is a transistor held at a chosen point on its load line while a small signal pushes it about. The point has to be chosen, which is a design problem, and it has to stay chosen, which is a rather harder one, because the most obvious way of setting it hands the choice straight back to beta.

Everything else here follows from getting that right. Once the operating point is nailed down you can ask what the signal does to it, and the answer turns out to be more interesting than the textbook formula suggests: the output of a common emitter stage is visibly the wrong shape long before anything is clipped. Then you can ask where the gain comes from, discover that most of it is not usable, and give a good deal of it away on purpose.

The circuit throughout is one stage: a supply, four resistors and a transistor. It is the smallest amplifier there is, and every large one is built out of stages that look very like it.

The load line, and the point you get to choose

A load line is a statement about the circuit outside the transistor. Put a supply and a resistance in series with the collector and emitter, and Kirchhoff says

V_CC = I_C(R_C + R_E) + V_CE

which is a straight line on the output characteristic, running from V_CE = V_CC at zero current up to I_C = V_CC/(R_C + R_E) at zero volts. The transistor must sit somewhere on that line, because the line is just the resistors and the supply, and they are not negotiable.

Where on it is decided by the base. That is the whole of biasing: the resistors draw the line, and the base picks the point. The point is called the quiescent point, or Q-point, and quiescent means what the circuit does when nothing is happening.

There are only three interesting places to put it. At the top of the line the transistor is saturated and behaves as a closed switch, which is what you want in the digital half of the module and useless here. At the bottom it is cut off and behaves as an open one. Somewhere in the middle it is in the active region, where a small change at the base produces a large and proportional change at the collector, and a stage biased there can go up as well as down.

The middle is not a comfortable place to be. It is the only one of the three where the transistor is dissipating real power, and it is the only one where the exact value matters, because a point set by a device parameter you do not control is not really a point you have set.

How to use it

Setting the operating point is the design problem. Choose the supply, the divider, the collector and emitter resistors, and read the Q-point off the schematic and the tables. The strip along the bottom solves the same circuit three times with the three different transistors, so the question “does this survive being handed a different part” is answered as a number rather than as an opinion. Switch the bias arrangement to the single base resistor and watch that number fall apart.

The signal on the bias adds a sine wave to the base and solves the whole circuit at a hundred and twenty points around the cycle. Nothing is multiplied by a gain figure, so the shape you see is the shape the circuit produces. The phase slider walks you round the cycle with a cursor on both waveforms and a dot sliding along the load line.

Where the gain comes from is the trade. It computes r_e from the operating current, works out the gain from it, and then lets you leave part of the emitter resistor unbypassed. The gain falls along a curve, and three other numbers improve as it does, including a distortion figure measured off the same solve rather than quoted.

Coupling and bypass puts the three capacitors in and plots the frequency response. Each one sets a corner, the corners are marked where they fall, and the audio band is shaded so that “too small” means something. The operating point is shown alongside and does not move, which is the entire reason the capacitors are there.

The three configurations computes the common emitter, common collector and common base stages from the same part at the same operating point, and puts them in one table. Drag the load resistance down towards a hundred ohms and watch which of the three still works.

Amplifier Stages, and Choosing the Operating Point

Biasing a stage so that beta stops mattering, the signal riding on the bias, where the gain comes from and what it costs, and the capacitors that keep the two apart.

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An amplifier is a transistor held at a chosen point on its load line.

A signal is a small disturbance. It has to disturb the circuit from somewhere, and that somewhere is the quiescent point, or Q-point.
Put it too high and the transistor saturates on the first loud passage. Too low and it cuts off. In the middle it has room to move both ways.
The difficulty is holding it there. The collector current follows beta, and beta is 60 on one of the parts below and 290 on another.
The fix is an emitter resistor and a divider stiff enough to hold the base voltage. Watch the three-part comparison at the bottom as you change things.

Transistor

Bias arrangement

Supply, Vcc12.0 V

R1, base to supply47 kΩ

R2, base to ground10 kΩ

Collector resistor, Rc3.3 kΩ

Emitter resistor, Re1 kΩ

Try one of these

Active, with room to move in both directions

V_CE sits at 6.19 V, which is 52 per cent of the supply.
The collector can rise 4.45 V and fall 5.99 V before it runs into something.
The stage is dissipating 8.4 mW in the transistor while doing nothing at all, which is what a class A stage costs.

What the base sees

V_TH2.11 V

R_TH8.2 kΩ

V_BE672.9 mV

I_B8.99 µA

divider / base23 : 1

The operating point

I_C1.35 mA

V_E1.36 V

V_C7.55 V

V_CE6.19 V

RegionActive

Room to move

upwards4.45 V

downwards5.99 V

usable peak4.45 V

P in the device8.4 mW

P from the supply18.7 mW

The same circuit, the same resistors, three different transistors

2N3904β 150

I_C1.35 mA

V_CE6.19 V

RegionActive

BC547β 290

I_C1.40 mA

V_CE5.99 V

RegionActive

BD139β 60

I_C1.30 mA

V_CE6.40 V

RegionActive

Beta runs from 60 to 290 across these three, a factor of nearly five.
The collector current here varies by only 1.08 to 1 and V_CE moves by 0.41 V, so the operating point is being set by the resistors.

Why the emitter resistor does that

Suppose the collector current tries to rise, because you fitted a higher gain part or the room got warmer.
A larger current through R_E raises the emitter voltage.
The base voltage is held where the divider put it, so V_BE, which is V_B minus V_E, gets smaller.
A smaller V_BE asks for less current, and the rise is cancelled before it happens. That is negative feedback, and this is the first circuit in the module that uses it.
It works only while the base voltage really is held. A divider passing ten times the base current holds it; one passing about the same as the base current does not, which is what the limp divider preset shows.

Walkthrough

Step 1: Read the operating point off the textbook stage

Open Setting the operating point and press The textbook stage: 12 V, a 47 kΩ and 10 kΩ divider, 3.3 kΩ in the collector and 1 kΩ in the emitter.

Work down the drawing. The divider would sit at 2.105 V with nothing attached, and that is V_TH. The base actually sits at 2.031 V, slightly lower, because the base is drawing current through the divider. Subtract V_BE, which the demonstration works out to be 0.673 V rather than assuming 0.7 V, and the emitter is at 1.358 V. That across 1 kΩ is 1.358 mA, and near enough all of it comes out of the collector, where it drops 4.45 V across 3.3 kΩ and leaves the collector at 7.55 V.

V_CE is 6.19 V, just over half the supply. The transistor can rise 4.45 V before the collector reaches the rail and fall 5.99 V before it saturates, so the stage has room to move in both directions.

Notice the last line of the third card: 8.4 mW dissipated in the transistor while the circuit is doing nothing at all. That is the standing cost of a class A stage, and it is paid whether there is a signal or not.

Step 2: Swap the transistor and watch nothing happen

Look at the strip along the bottom, which solves the same four resistors with all three parts.

The BD139 has a beta of 60, the 2N3904 has 150 and the BC547 has 290, a spread of nearly five to one. The collector currents are 1.298 mA, 1.349 mA and 1.397 mA, a spread of 1.08 to one. V_CE moves by 0.41 V across the whole range.

That is the result the arrangement exists to produce. The operating point is being set by four resistors and is barely aware of which transistor is in the socket.

Step 3: Take the emitter resistor away

Press No emitter resistor. The divider has been made much weaker to compensate, so the 2N3904 still lands in a sensible place at 1.46 mA and 7.18 V.

Now look at the strip again. The three currents are 0.70 mA, 1.46 mA and 2.78 mA, a spread of very nearly four to one, and V_CE ranges over 6.86 V of a 12 V supply. The BC547 is close to saturating and the BD139 is close to cutting off.

This is the step worth being clear about, because it is the one that catches people. The divider is not what stabilises the bias. A divider on its own sets a base voltage, and a base voltage on its own sets nothing useful, because the current that voltage produces is exponential in it and multiplied by beta. It is the emitter resistor that does the work, and the divider is there to make the emitter resistor able to do it.

Step 4: Try the arrangement that does not work

Put the emitter resistor back with The textbook stage, then switch Bias arrangement to Single base resistor.

The default 1.1 MΩ is chosen so that the 2N3904 lands on almost exactly the Q-point it had before: 1.358 mA and 6.15 V. On this part, the two arrangements are indistinguishable.

Now read the strip. The BD139 sits at 0.59 mA and 9.45 V, and the BC547 at 2.36 mA and 1.84 V. The current spread is four to one and V_CE ranges over 7.6 V. One part is halfway to cutting off and the other is halfway to saturating, and neither of them will amplify properly.

The reason is visible in the base loop. With a single resistor from the supply, the base current is set by that resistor and by nothing else, so the collector current is beta times a fixed number. Beta changes, so the collector current changes with it. The divider and emitter resistor arrangement never lets the base current decide anything: it fixes a voltage and lets the emitter resistor turn that voltage into a current.

Step 5: Watch the divider go limp

Go back to the divider and press A limp divider, which keeps the same 2.105 V ratio but uses 470 kΩ and 100 kΩ instead of 47 kΩ and 10 kΩ.

The base voltage has fallen from 2.031 V to 1.596 V, the collector current from 1.349 mA to 0.927 mA, and the three-part spread has opened out from 1.08 to 1.77. The stiffness readout in the first card has dropped from 23:1 to 3:1.

That readout is the ratio of the current running through the divider to the current the base takes out of it. The argument in the note above assumed the base voltage is held, and a divider passing only three times the base current does not hold anything: when beta changes, the base current changes, and the divider moves with it.

Then press A stiffer divider, which is 4.7 kΩ and 1 kΩ. The stiffness is now 223:1 and the spread is down to 1.03. It also wastes 2.1 mA continuously in the divider, which is more than the stage itself draws, so ten to one is the usual compromise rather than the largest number you can afford.

Step 6: Find out what “small signal” actually means

Open The signal on the bias and press A genuinely small signal, which is 2 mV peak.

The output is a clean sine wave, upside down, swinging about 0.34 V either side of the quiescent 7.55 V. The measured gain is about 172, which is what the theory says it should be. The two half cycles differ by about 8 per cent.

Now press Already distorting, which is 10 mV. Still nothing is clipped, nothing has hit a rail, and the output is visibly lopsided: 2.10 V down against 1.43 V up, a mismatch of 38 per cent.

This is the point of the tab. The signal is 10 mV against a thermal voltage of 26 mV, and that ratio is what decides the distortion. A junction gives more extra current for a few millivolts more than it gives up for a few millivolts less, because the relationship is exponential, so the downward half of the output is always the bigger one. Small signal means small compared with 26 mV, not small compared with the supply.

Step 7: Push it until the bottom goes flat

Press Hard into saturation, which is 40 mV.

The collector now spends 37 of the 121 samples pinned at 1.56 V. That is V_E plus V_CE(sat), the transistor is fully on, and the output has a flat bottom with a corner on it. Walk the phase slider through the positive half of the input and watch the cyan dot on the load line run down to the wall and stop.

Then look at the top of the same waveform. It is squashed, but it is rounded rather than flat, and it never quite reaches the supply.

The two ends fail differently and it is worth knowing why. At the bottom the transistor runs out of transistor: V_CE cannot go below V_CE(sat) and the device stops dead. At the top it is turning off, and turning off is something an exponential does gradually. Halving the base drive does not halve the current, it divides it by about four, so the collector creeps towards the supply instead of arriving at it.

Step 8: Buy linearity with gain

Press Bypass removed, still at 40 mV. The clipping is gone, the distortion has fallen to a tenth of a per cent, and the output is 0.26 V peak to peak rather than 9.5. The gain has collapsed from about 170 to 3.2.

Open Where the gain comes from to see what happened.

The stage runs at 1.36 mA, so r_e, which is V_T divided by I_E, is 19 Ω. With the emitter fully bypassed the signal sees only that 19 Ω, and the gain is 3.3 kΩ divided by 19 Ω, which is 173. With the bypass removed it sees 19 Ω plus the whole 1 kΩ resistor, and the gain is 3.3 kΩ divided by 1019 Ω, which is 3.24.

Now drag Unbypassed emitter, Re1 slowly up from zero and watch four numbers at once:

R_E1	Gain	Input resistance	Half-cycle mismatch
none	−173	2.1 kΩ	15.4 %
22 Ω	−80	3.5 kΩ	3.3 %
47 Ω	−50	4.5 kΩ	1.3 %
100 Ω	−28	5.6 kΩ	0.40 %
300 Ω	−10	7.0 kΩ	0.06 %

Forty-seven ohms costs you two thirds of the gain and takes the distortion down by more than a factor of ten. That is the bargain, and it is the one nearly every real amplifier makes, because you always have more gain than you need and never more linearity than you need.

There is a fifth thing that improves and the readouts cannot show it. Once R_E1 is much bigger than r_e, the gain is R_C divided by R_E1, and neither of those is the transistor. The gain has stopped depending on the operating current, on the temperature and on which part is in the socket, and has become a ratio of two resistors you chose.

Step 9: Put the capacitors in

Open Coupling and bypass. This is the same stage with a source on the front, a load on the back and three capacitors.

The two coupling capacitors are there so that the source cannot drag the base voltage away from where the divider put it, and so that the load cannot drag the collector. Change any of the three sliders and check the right hand card: the operating point does not move at all. That is what coupling means.

What the capacitors do cost you is the bottom of the frequency range. Each one sits in series with the resistance it faces, so each is a high pass filter, and the graph marks the three corners where they fall. With the defaults they are at 5.8 Hz, 1.2 Hz and 8.5 Hz, all comfortably below the shaded audio band.

Then look at the values that produced them. C_in and C_out are 10 µF and the bypass capacitor is 1000 µF, a hundred times larger. Press Bypass too small only to see what happens when it is not: the response develops a step, falling from 40 dB at the top of the band down towards 5.6 dB at the bottom, which is the gain of the same stage with the emitter resistor unbypassed.

Finally press A high resistance source. Nothing about the shape of the response changes, but the whole curve drops by 13 dB. The source has 10 kΩ of its own resistance and the stage offers 2.1 kΩ, so most of the signal is lost in the divider they form before the amplifier sees any of it. No amount of gain afterwards recovers it, which is the argument for the last tab.

Step 10: Compare the three configurations

Open The three configurations. The same transistor at much the same operating point, wired three ways.

At a 10 kΩ load the table reads: the common emitter gives a gain of −130 with an input resistance of 2.1 kΩ; the common collector gives 0.98 with an input resistance of 7.8 kΩ and an output resistance of 69 Ω; the common base gives +130 with an input resistance of 18.7 Ω.

Now drag the load down to 470 Ω. The common emitter collapses to −22, because the load is in parallel with R_C and is now much the smaller of the two. The follower goes from 0.98 to 0.94 and carries on as if nothing had happened.

That is what a follower is for. It has no voltage gain to lose, so a heavy load costs it almost nothing, and it converts a signal that exists at high resistance into the same signal at 69 Ω. The standard arrangement is one of each: a common emitter stage to get the gain, and a follower behind it to drive whatever comes next.

The common base stage looks like the common emitter with the sign changed, and the number that actually distinguishes it is the input resistance: 18.7 Ω against 2.1 kΩ. That is a liability almost everywhere and an advantage in exactly two places, which are the end of a coaxial cable and the front of a radio frequency stage.

Check your understanding

Quiz

Select 0/1

The same divider and the same four resistors, and the transistor is swapped from one with a β of 60 to one with a β of 290. What happens to the collector current?

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Quiz

Select 0/1

Why does adding an emitter resistor make the operating point insensitive to β?

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Concept Match

Match each part of the stage to the job it does

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Quiz

Select 0/1

A stage runs at 1.35 mA with a 3.3 kΩ collector resistor, a 1 kΩ emitter resistor and the emitter fully bypassed. Roughly what is the voltage gain?

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Quiz

Select 0/1

An amplifier with an output resistance of 3.3 kΩ has to drive a 100 Ω loudspeaker. What should go between them?

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Wrap-up

Six things to take away.

The resistors draw the load line and the base picks the point on it. That is the whole of biasing. Both halves have to be deliberate, and the second one is the harder of the two.
A divider alone does not stabilise a bias. Take the emitter resistor out and the collector current spreads four to one across three ordinary transistors, whatever the divider is doing. The emitter resistor is the mechanism, and the divider exists to make it work.
Negative feedback is the idea underneath all of it. A rise in current is subtracted from V_BE and cancelled. The same idea, applied to the signal rather than the operating point, is what an unbypassed emitter resistor does.
Small signal means small compared with 26 mV. Ten millivolts at the base of an unbypassed stage produces a visibly lopsided output with nothing clipped anywhere. The thermal voltage, and not the supply, is the yardstick for whether a signal is small.
Gain is the currency you spend on everything else. Forty-seven ohms in the emitter costs two thirds of the gain and buys a tenfold reduction in distortion, twice the input resistance, and a gain that no longer depends on the transistor at all. You always have more gain than you need.
The three configurations are three different jobs. Common emitter for voltage gain, common collector to drive something heavy from something weak, common base for a low input resistance. The usual arrangement is a common emitter stage followed by a follower, and between them they do what neither can do alone.

The idea worth carrying furthest is the fourth one, because it generalises. Every device in this module has been described by a curve, and every curve has been approximated by a straight line somewhere. The approximation is only ever good over a small enough range, and this is the first demonstration where you can see exactly how small “small enough” has to be, and what it costs to make the range bigger.